Sol. We will first determine the deviation produced by the system. From the figure, we see that

n sin α = n′ sin θ ′,

n′ sin θ ′′ = 1 . sin θ ′′′.

Now,

deviation of ray δ = ( θ ′ – α ) + ( θ ′′′ – θ ′′)

From triangle QRB, we have

α + 90º + θ ′′ + 90º – θ ′ = 180º

⇒ θ ′′ = θ ′ – α .

Since angle α is small, θ ′, θ ′′, θ ′′′ are also small.

∴ n α = n′ θ ′′ and n′ θ ′′ = θ ′′′.

Solving θ ′, θ ′′ and θ ′′′ in terms of α , θ ′ = α ,

θ ′′ = α – α = α and θ ′′′ = n′ α = (n – n′) α .

∴ δ = α – α + (n – n′) α – = (n – n′) α

If the source is at a distance a , then

d (distance between virtual sources) = 2a δ = 2a(n – n′) α

D = a + b where b = distance of screen from the system.

∴ β = =

For plane wave α → ∞ ,

∴ β = = = 0.284 mm.